I have two questions:
If $K$ is a field, $R=K[x_1,\ldots,x_n]$, the ring of polynomials over $K$ with $n$ indeterminates, and $M$ is a maximal ideal of $R$ why is the contraction $N$ of $M$ to $K[x_1,\ldots,x_{n-1}]$ maximal?
Why $M$ lies properly above $NR$?
Thanks!
Let $S=K[X_1,\dots,X_{n-1}]$. Now consider the polynomial extension $S\subset S[X]$. The question becomes
Why a maximal ideal of $S[X]$ lies over a maximal ideal of $S$?
This holds in the more general frame of finitely generated algebras over a field, and a proof can be found here. (However, one can not extend the property too much since even for $S$ a noetherian UFD this is not true; for a counterexample see here.)
In general, if $N=M\cap S$ then it's easy to see that $NS[X]\subsetneq M$ for the simple reason that $NS[X]$ is not maximal ($S[X]/NS[X]\simeq (S/N)[X]$).
